A 28.0-kg block is connected to an empty 2.00-kg bucket by a cord running over a frictionless pulley (Fig. 4–73). The coefficient of static friction between the table and the block is 0.45 and the coefficient of kinetic friction between the table and the block is 0.32. Sand is gradually added to the bucket until the system just begins to move. (a) Calculate the mass of sand added to the bucket. (b) Calculate the acceleration of the system. Ignore mass of cord.

Draw the free body diagram for the bucket and block, separately, and mention all the forces acting on it, such as frictional force, the weight of the block and others. When the block reaches its maximum position, its static frictional force increases.

Given data:

The mass of the block is\(m = 28\;{\rm{kg}}\).

The mass of the bucket is\(m' = 2\;{\rm{kg}}\).

The static coefficient of friction between the table and the block is\({\mu _s} = 0.45\).

The kinetic coefficient of friction between the table and the block is\({\mu _k} = 0.32\).

The free body diagram of the block is as follows:

The free body diagram of the bucket is as follows:

The relation to calculate the forces in the y-direction of the bucket is given by:

\(\begin{aligned}\Sigma {F_{yb}} &= 0\\{m_1}g - {F_T} &= 0\\{F_T} &= {m_1}g\end{aligned}\)

Here, \({F_T}\) is the normal force on the bucket, \({m_1}\) is the total mass, and g is the gravitational acceleration.

The relation to calculate the forces in the x-direction of the block is given by:

\(\begin{aligned}\Sigma {F_x} &= 0\\{F_T} - {F_{fr}} &= 0\\{F_T} &= {F_{fr}}\end{aligned}\)

Here,\({F_{fr}}\)is the frictional force exerting on the block.

On plugging the values in the above relation, you get:

\(\begin{aligned}{m_1}g &= {\mu _s}{F_N}\\{m_1}g &= {\mu _s}mg\\{m_1} &= \left( {0.45} \right)\left( {28\;{\rm{kg}}} \right)\\{m_1} &= 12.6\;{\rm{kg}}\end{aligned}\)

The amount of mass added is calculated as:

\(\begin{aligned}{m_a} &= {m_1} - m\\{m_a} &= \left( {12.6\;{\rm{kg}}} \right) - \left( {2\;{\rm{kg}}} \right)\\{m_a} &= 10.6\;{\rm{kg}}\end{aligned}\)

Thus, \({m_a} = 10.6\;{\rm{kg}}\) is the mass of the sand added.

The relation to calculate the forces in the y-direction of the bucket is given by:

\(\begin{aligned}\Sigma {F_y} &= {m_1}a\\{m_1}g - {F_T} &= {m_1}a\\{F_T} &= {m_1}g - {m_1}a\end{aligned}\)

The relation to calculate the forces in the x-direction of the block is given by:

\(\begin{aligned}\Sigma {F_x} &= ma\\{F_T} - {F_{fr}} &= ma\\{F_T} &= {F_{fr}} + ma\end{aligned}\)

On equating the above two relations of tension force, you get:

\(\begin{aligned}{m_1}g - {m_1}a &= {F_{fr}} + ma\\{m_1}g - {m_1}a &= {\mu _k}{m_2}g + ma\\a &= \left( {\frac{{{m_1} - {\mu _k}m}}{{{m_1} + m}}} \right)g\end{aligned}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}a &= \left( {\frac{{12.6\;{\rm{kg}} - \left( {0.32} \right)\left( {28\;{\rm{kg}}} \right)}}{{\left( {12.6\;{\rm{kg}}} \right) + \left( {28\;{\rm{kg}}} \right)}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\\a &= 0.87\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

Thus, \(a = 0.87\;{\rm{m/}}{{\rm{s}}^2}\) is the acceleration of the system.